Question

Find the zeros of y=x^2 - 4x - 9 by completing the square

Answer 1

`To find the zeros of the quadratic equation \(y = x^2 - 4x - 9\) by completing the square, follow these steps:`

`1. Begin with the equation: \(y = x^2 - 4x - 9\).`

`2. Group the \(x^2\) and \(x\) terms together: \[y = (x^2 - 4x) - 9\]3. To complete the square for the quadratic term \(x^2 - 4x\), take half of the coefficient of the \(x\) term (-4), square it, and add and subtract that value within the parentheses:`

`\[y = (x^2 - 4x + (-4)^2) - (4^2) - 9\] \[y = (x^2 - 4x + 4) - 16 - 9\]4. Simplify the equation: \[y = (x - 2)^2 - 25\]Now, the equation is in the form \(y = (x - h)^2 - k\), which represents a parabola with the vertex at the point \((h, k)\).`

`The zeros occur when \(y = 0\):\[0 = (x - 2)^2 - 25\]Adding 25 to both sides:\[25 = (x - 2)^2\]Taking the square root of both sides:\[√(25) = √((x - 2)^2)\]\[5 = |x - 2|\]`

Now, solve for \(x - 2\) by considering both the positive and negative values of 5:

`\(x - 2 = 5\) or \(x - 2 = -5\)Solving each equation for \(x\):\(x = 5 + 2 = 7\) or \(x = -5 + 2 = -3\)Therefore, the zeros of the equation \(y = x^2 - 4x - 9\) obtained by completing the square are \(x = 7\) and \(x = -3\).`